Problem: Find $\lim_{h\to 0}\dfrac{4\sin\left(\dfrac{\pi}{2}+h\right)-4\sin\left(\dfrac{\pi}{2}\right)}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $4$ (Choice D) D The limit doesn't exist
Solution: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $4\sin\left(\dfrac{\pi}{2}+h\right)-4\sin\left(\dfrac{\pi}{2}\right)$, we can tell that the function is $f(x)=4\sin(x)$ and the $x$ -value is $\dfrac{\pi}{2}$. In other words, the limit expression is equal to $f'\left(\dfrac{\pi}{2}\right)$ for $f(x)=4\sin(x)$. Let's find $f'(x)$ : $f'(x)=4\cos(x)$ Now let's evaluate $f'\left(\dfrac{\pi}{2}\right)$ : $f'\left(\dfrac{\pi}{2}\right)=4\cos\left(\dfrac{\pi}{2}\right)=0$ In conclusion, $\lim_{h\to 0}\dfrac{4\sin\left(\dfrac{\pi}{2}+h\right)-4\sin\left(\dfrac{\pi}{2}\right)}{h}=0$.